of 1 64

Proposal For A Characteristic Length Associated With Habitable Star Systems And For Life As

A Natural Universal Process!

By Ian Beardsley!

of 2 64

Contents

Abstract………………………………………………………………………….3

Part 1: Earth/Moon/Sun System And The Proton………………………..4!

Important To Point Out………………………………………………………………………5

Introduction………………………………………………………………………6!

Earth/Moon/Sun System……………………………………………………….6!

Proton Radius……………………………………………………………………………………10

Part 2: The Solar Formulation……………………………………………..13

Equating The Lunar And Solar Formulations Yield

Our 1 Second Base Unit………………………………………………………………………17

Part 3: The Characteristic Length For Habitable Planets………..22

Part 4: The Ancient System of Measuring Time

and the Contemporary Second…………………………………………….25

Part 5: Biological Life Part of a Universal

Natural Process In The Universe………………………………………….35

Part 6: Solutions For Jupiter And Saturn……………………………..42

Part 7: Modeling Habitable Star Systems………………………………47

Part 8: The Meter A Natural Unit……………………………………………53

Appendix 1…………………………………………………………………………61

Appendix 2: The Data For Verifying The Equations…………………….63

of 3 64

Abstract!

By ﬁnding the solution to the Solar System of the Schrödinger Wave Equation used to solve the

atom in quantum mechanics, a characteristic length for habitable planets of star systems is

proposed. It gives the orbit of the Moon which we ﬁnd is is proportional to the cube of the

planet’s mass to the cube of its Moon. The moon is included because our moon makes life

optimally possible because its orbit holds the Earth at its inclination to its orbit allowing for the

seasons, preventing cold and hot temperature extremes. Part of the solution involves the

property of our Earth/Moon/Sun system that it has a near perfect eclipse which is to say that

the the Moon as seen from the Earth near perfectly eclipses the Sun.!

The solution exists in the formation of spacetime operators that have a characteristic time of 1

second. This is the way it happens and is interesting because the second precedes the

development of modern physics and ultimately has its origins in the ancient Sumerians, who

were of the ﬁrst to settle down from wandering and gathering to invent agriculture, writing, and

mathematics — civilization in general. Our unit of a second stems from their base 60 counting

system. So that is looked at and it may be that the dynamics of base 60 are responsible for the

naturalness of the second. The spacetime operators also solve the proton that is the

fundamental unit of charge for atoms.!

It is being suggested that life might be part of a Universal Natural Process, not just because of

what we just said above, but because the same spacetime operators that solve the Solar

System and the atom also are based on a sixfold symmetry found in the hydrocarbons which

are are the skeletons of life chemistry.!

I think this theory presenting life as a natural universal process has value in that it gives us a

context to look at current times where we have sent men to the Moon and robotic landers to

the planet Mars. A time where we are planning manned mission to Mars, as well as making

plans to colonize the red planet. In that we are now an interplanetary spacefaring civilization,

perhaps to soon become one that is interstellar."

of 4 64

Part 1: Earth/Moon/Sun System And The Proton

of 5 64

Important To Point Out

You can speak of the structure of the solar system even though it changes with time. This is

important to understand when I refer to sizes of the Moon and the planets, and their orbital

distances.

The whole object of developing a theory for the way planetary systems form is that they meet the

following criterion: They predict the Titius-Bode rule for the distribution of the planets; the

distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The

distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from

the parent star, the outward pressure gradient from the stellar production of radiation, and the

outward inertial forces as a cloud collapses into a flat disc around the central star. These forces

separate the flat disc into rings, agglomerations of material, each ring from which a different

planet forms at its central distance from the star (it has a thickness). In a theory of planetary

formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of

planets today, which was the distribution of the rings from which the planets formed.

The data in this paper is in Appendix 1 at its end.

I would like to clarify something here about the Moon:

I have shown the second in the paper is the base unit for both the Earth/Moon/Sun system as a

wave solution and the proton, making them connected to things like G, c, and h and that we

have this second when things are such that there is a perfect eclipse. The theory does not depend

on the eclipse, just the Radius of the Sun to the Radius of the Moon for the wave solution which

is the orbital radius of the Earth to the orbital radius of the Moon which is the condition of a

near perfect eclipse.

We got the second from the rotation period of the Earth at the time the moon came to perfectly

eclipse the Sun.

That is as the Earth’s day gets longer and the lunar orbit grows larger, we got the second at the

time that the Earth day was what it is during the epoch when the Moon perfectly eclipses the

Sun, 24 hours.

The near perfect eclipse is a mystery in the sense that it came to happen when anatomically

modern humans arrived on the scene, even before that, perhaps around Homo Erectus and the

beginning of the Stone Age. The Earth day was 18 hours long, long before that, 1.4 billion years

ago. Homo Erectus is around two to three million years ago.

I wrote in the first paragraph at the bottom:

"...but the theory prompts one to suggest the perfect eclipse has perhaps been a message to

humans, since hunting with stone spearpoint to present day having gone to the Moon, to tell us

we are here for reason."

I didn't say it wasn't a coincidence, but that it "prompts one to suggest" It has no bearing on the

theory.

Also, the Earth has been in the habitable zone since 4 billion years ago (0.9 AU) and that

habitable zone can continue to 1.2 AU. So we can speak of the distance to the Earth over much

time. The Earth and Sun formed about 4.6 billion years ago.

of 6 64

Introduction !

The Natural constants, such as the constant of gravitation G, the Planck constant for the

atoms, h, and the speed of light c describe the properties of space and time. I have found they

are very conducive to our base unit for measuring time, the second. However, the base unit of

a second was given to us by the ancient Sumerians of Mesopotamia who were among the ﬁrst

to settle down from wandering and gathering, from hunting with stone spearpoints to create

agriculture, writing, and mathematics. They divided the day from sunrise to sunset into 12 units

that we call today in the West, hours. As such there are 24 hours in a day from sunrise to

sunrise. They had a base 60 counting system because 60 is evenly divisible by so much, like

1,2,3,4,5,6,…the ﬁrst six integers. It is the smallest number to do this. The Babylonians

adopted this base 60 (called sexagesimal) from the Sumerians, and divide their hour into 60

minutes, and the minutes into 60 seconds. We got our system from the Babylonians. Since one

day (24 hours) is the time for the Earth to make one revolution on its axis, the second comes

from dividing the Earths rotation by these numbers. But I didn’t just ﬁnd that the natural

constants described the second, but that they formed spacetime operators that solve the atom

today in modern physics, and the solar system with the Schrödinger wave equation that

describes the atom in quantum mechanics. I further found that this is done with the Earth’s

moon, which seems to be a natural yardstick for measuring size and mass. This says a great

deal about the mystery known to astronomy for a long time that the Moon perfectly eclipses

the Sun as seen from the Earth. In fact, it becomes part of the solution to planetary systems,

like our solar system, for them to have life optimally possible. We know the Moon orbiting the

Earth optimizes the condition for life on Earth by holding the Earth at its tilt to its orbit around

the Sun allowing for the seasons and thus preventing extreme cold and extreme hot. Not only

does the lunar eclipse like this play a key role in the equations, but the theory prompts one to

suggest the perfect eclipse has perhaps been a message to humans, since hunting with stone

spearpoint to present day having gone to the Moon, to tell us we are here for reason.!

Earth/Moon/Sun System This second I found is the base unit of the orbital dynamics of the

solar system. I found I could create three spacetime operators, one that acts on the radius of

the proton to its mass, and the other that acts on the radius of the Sun to its mass, and one

that acts on angular momentum to speed. is Planck’s constant, is the universal constant of

gravitation, is the speed of light, is the ﬁne structure constant, is the radius of a proton,

is the mass of a proton, is the mass of the Moon, is its radius, and the EarthDay is

the rotation period of the Earth:!

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

of 7 64

They seem to suggest that the Earth orbit might by quantized in terms of the Moon and a base

unit of one second, which we will see can be considered a characteristic time of the Universe.

Equation 2 is actually 1.3 seconds but rounds to one second. However it is derived from another

equation that gives 1.2 seconds. Namely

It says one second is the Earth Day (completion of one rotation of the Earth) adjusted by the

kinetic energy of the Moon, to the kinetic energy of the Earth. As we shall see, the quantization

in terms of the Moon is exactly 1 second to nearly two places after the decimal from the ground

state of our solution to the Earth/Moon/Sun system.

We say there is a Planck constant for the solar system. We do it with a base unit of one second

because equations 1, 2, and 3 suggest we should . We suggest it is such that it is given by the

standard Planck constant for the atom, , times some constant, , and the kinetic energy of the

Earth.

Where

6).

Where equation 6 comes from equation 3.

We derive the value of our solar Planck constant

1secon d =

K E

(Ear th Da y)

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

π r

G m

C =

⋅

2π r

G m

⋅

18769

299792458

⋅

2π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

1.55976565E 33

⋅

kg ⋅ m

⋅

= kg

= energ y

hC = (6.62607E − 34)(1.55976565E33) = 1.03351secon ds ≈ 1.0secon d s

of 8 64

Now we show that our Planck constant for the solar system gives the base unit of one second for

the quantization. We call our solar Planck constant and find the wavelength for the Moon

which is the ground state for the solar system:

The wavelength associated with the Moon divided by the speed of light should be 1 second if our

planetary system is quantized in terms of the Moon and a base unit of one second. We have

And we see it is, so we have

The solution for the orbit of the Earth around Sun with the Schrödinger wave equation can be

inferred from the solution for an electron around a proton in the a hydrogen atom with the

Schrödinger wave equation. The Schrödinger wave equation is, in spherical coordinates

10)

Its solution for the atom is as guessed by Niels Bohr before the wave equation existed:

11)

12)

hC =

(

⋅ m ⋅ s

)

(

⋅

)

(

)(

⋅

)

= secon d s

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

ℏ

⊙

= (hC )K E

earth

= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s

ℏ

⊙

moon

ℏ

⊙

(2.8314E 33)

(6.67408E − 11)(7.34763E 22kg)

= 3.0281E8m

moon

3.0281E8m

299,792,458m /s

= 1.010secon d s

moon

= 1secon d

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

E = −

)

2ℏ

ℏ

Z k

of 9 64

is the energy for an electron orbiting protons and , is the orbital shell for an electron with

protons, the orbital number. I find the solution for the Earth around the Sun utilizes the

Moon around the Earth. This is different than with the atom because planets and moons are not

all the same size and mass like electrons and protons are, and they don’t jump from orbit to

orbit like electrons do. I find that for the Earth around the Sun

13)

14)

is the kinetic energy of the Earth, and is the planet’s orbit. is the radius of the Sun,

is the radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is

the orbit number of the Earth which is 3 and is the Planck constant for the solar system.

Instead of having protons, we have the radius of the Sun normalized by the radius of

the Moon. It is has always been an amazing fact the the sizes of the Moon and the Sun are such

that given their orbital distances, the Moon as seen from the Earth perfectly eclipses the Sun.

This becomes part of the theory and we suggest it is a condition for sophisticated planets that

harbor life by writing it:

15)

That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is

about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say

the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon

orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its

inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and

extreme cold. Let us now see if our solar Planck constant works…

16)

=2.727E36J

The kinetic energy of the Earth is

The accuracy of our equation is:

K E

= n

⊙

⋅

2ℏ

⊙

2ℏ

⊙

⋅

⊙

⋅

K E

⊙

ℏ

⊙

planet

moon

≈

star

moon

⊙

6.96E8m

1737400m

= 400.5986

K E

= (1.732)(400.5986)

(6.67408E − 11)

(5.972E 24kg)

(7.347673E 22kg)

2(2.8314E 33)

K E

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

2.727E 36J

2.7396E 33J

100 = 99.5 %

of 10 64

Which is very good. Thus we have solved the Earth/Moon/Sun System with our spacetime

operators by using them to find a Planck constant for the solar system.

We want to suggest that the base 60 dynamic applied to the rotational velocity of the Earth to

measure time has a natural property associated with the atom and the Earth/Moon/Sun orbital

system as a solution to the atom’s wave equation. The Earth day is given by, in seconds

This gives since

Thus we have as factors for the seconds per day the smallest primes 2 and 3, and the 4 of

rectangular coordinate systems, and the versatile, abundant, 60. We then suggest the second is

dynamic in terms of what the ancients gave us because 86400 seconds comes from

and this I suggest is connected to Nature because the Earth rotates at a speed

that gives from these ancient factors the duration of a second that I found is the base unit of the

atomic systems and our planetary system, in particular of our Earth/Moon/Sun system as a

solution of the Schrödinger wave equation that solves the same atomic system, which I will go

into right now.

Proton Radius We see the spacetime operators solve the atom by giving us the radius of a

proton. We set equation 3 equal to equation 1

1).

These two yield

17).

I find this is close to the experimental value of the radius of a proton. I find I can arrive at this

radius of a proton another way, energy is given by Planck’s constant and frequency

We have

(

1d a y

24hrs

)(

1hr

60mi n

)(

1min

60sec

)

86400

d a y

sec

2 ⋅ 3 ⋅ 4 = 24

2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60 = 86400secon d s /day

2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60

⋅

π r

G m

⋅

= 1secon d

(

6α

4πh

)

⋅

= 1secon d

⋅

E = h f

f = 1/s, h = J ⋅ s, h f = (J ⋅ s)(1/s) = J

E = J = Joules = en erg y

of 11 64

We take the rest energy of the mass of a proton :

The frequency of a proton is

Since our theory gave us the factor of 2/3 for the radius of a proton we have:

The radius of a proton is then

This is very close to the value upon which the proton radius converged historically by two

independent methods which was 0.877E-15m. The result from our theory is

The 0.877fm was challenged in 2010 by a third experiment making it 4% smaller and was

0.842E-15m. We find it may be that the radius of a proton is actually

18)

Where is the golden ratio constant (0.618…). This is more along the lines of more recent

measurements. Both equations 17 and 18 for the radius of the proton can be right depending on

the dynamics of what is going on; the radius of a proton is not precisely defined, it is more of a

fuzzy cloud of subatomic particles. Thus we have solved the atom with our spacetime operators

by producing the radius of a proton. I began working on this theory when the proton radius was

0.833fm so it is what I have been using in this paper. We continue to be honing in on its

experimental value every year with more experiments. The vast gap between the historical

0.877fm and the 2010 0.842fm is known in physics as the proton puzzle.

Indeed over the full range of values for the radius of a proton, the characteristic time is very

close to one second. The large value they got a long time ago gives:

E = m

⋅

≈ ϕ =

⋅

6.62607E − 34

(299,792,458)(1.67262E − 27)

= 0.88094E − 15m

= ϕ ⋅

= 0.816632E − 15m

of 12 64

The current value, which they believe they have really closed in on as very accurate in two

different experiments in 2019 using different methods, which is , gives a

time of 1.00500 seconds.

(

6α

4πh

)

⋅

= 1secon d

(18769)((0.877E − 15m)

6(1.67262E − 27kg)

(6.62607E − 34J ⋅ s)(4π)

(6.674E − 11)(299,792,458m /s

= 1.0581secon d s

= 0.833E − 15m

of 13 64

Part 2: The Solar Formulation

of 14 64

The Solar Formulation

Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the

mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the

Sun. In our lunar formulation we had:

We remember the Moon perfectly eclipses the Sun which is to say

Thus equation 1 becomes

The kinetic energy of the Earth is

Putting this in equation 3 gives the mass of the Sun:

We recognize that the orbital velocity of the Moon is

So equation 5 becomes

This gives the mass of the Moon is

Putting this in equation 1 yields

K E

= 3

⊙

⋅

2ℏ

⊙

K E

= 3

⋅

2ℏ

⊙

K E

⋅

⊙

= 3r

⋅

ℏ

⊙

= 3r

⋅ v

⋅

ℏ

⊙

ℏ

⊙

K E

⊙

⋅

⊙

of 15 64

We now multiply through by and we have

10.

Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular

momentum quantized, the angular momentum we will call , the subscript for Planck. We

have

We write for the solution of the Earth/Sun system:

11.

Let us compare this to that of an atom:

12.

We notice that in equation 11

; ; ; ;

is really . We can write 11 as

13.

We say. . That is

Let us see how accurate our equation is:

K E

⊙

⋅

⊙

ℏ

⊙

= r

= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg ⋅

= r

= 7.4483E 77J ⋅ m

⋅ kg = 8.3367E 77kg

⋅

K E

⊙

⋅

⊙

E = −

⋅

2ℏ

→

⊙

→ G

→ M

⊙

ℏ

→ L

ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

= ℏ

⊙

/2π

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

⊙

⋅

⊙

2ℏ

⊙

of 16 64

We have that the kinetic energy of the Earth is

Our equation has an accuracy of

Which is very good.

We call the same in our solar solution. But we now want our solution for the solar

formulation. is the mass of the Earth and is the mass of the Sun. We have our solution

might be!

14. !

Where is !

We have

15.

This has an accuracy of

Thus the solutions to the wave equation!

⊙

(6.67408E − 11)

(5.972E 24kg)

(1.9891E 30kg)

2(8.3367E 77kg

⋅

)

⊙

(6.759E 30J )

⊙

6.957E8m

1737400m

= 400.426

K E

= 2.70655E 33J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

2.70655E 33J

2.7396E 33J

= 98.79 %

ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

(9.13E 38)

(6.67408E − 11)(5.972E 24kg)

⋅

400.5986

= 1.4638E11m

1.4638E11m

1.496E11m

100 = 97.85 %

of 17 64

for the solar formulations are!

13)

14)

Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit

Let us equate equation 1 with equation 13:

13.

This gives:

16.

We remember that

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

= n

⊙

⋅

2ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

⋅

2ℏ

⊙

⋅

⊙

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

2π r

G m

of 18 64

This gives

17.

We have

18.

This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in

terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,

and the orbital number of the Earth. Let us compute

19.

Let us see how well equation 18 works. at aphelion is 966 m/s and .

. We have

That is an accuracy of

Equation 18 can be written:

20.

From our equation:

6α

h 4π r

= 1.004996352secon ds

6α

⋅

h 4π

⋅

⊙

(1secon d )

⊙

(5.972E 24kg)

(1.9891E 30kg)

(7.34763E 22kg)

(1.732)

= 321,331.459 ≈ 321,331

= 29,800m /s

= 1AU = 1.496E11m

2(966m /s) =

(29,800m /s)

1.496E11m

(1sec)(321,331.459)

1,907m /s = 1,932m /s

1907

1932

= 98.7 %

1secon d = 2r

⊙

6α

h 4π r

= 1.004996352secon ds

of 19 64

We have

21.

Since , the diameter of the Earth orbit, we have

22.

And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice

versa. We basically have

23. 62)

Where is the Earth orbital number. We have

= 7.83436E4seconds

EarthDay=(24)(60)(60)=86400 seconds,

accuracy

This last equation, equation 62, we can use to find the rotation period, or the length of the day,

of an earth-like planet in the habitable zone of any star system, so it would be very useful.

We want to turn our attention to equation 20 and write it

6α

⋅

h 4π

= 2r

⋅

⊙

= d

6α

⋅

h 4π

= d

⋅

⊙

1secon d = 2v

⊙

1secon d =

K E

moon

K E

earth

Ear th Da y

1secon d =

Ear th Da y

Ear th Da y =

⊙

n = 3

Ear th Da y =

2(1.496E11m)

966m /s

7.34763E 22kg

1.9891E 30kg

1.732

7.834E4s

86,400s

100 = 90.675 %

of 20 64

24.

We see equating solar and lunar formulations for energy yield the base unit of one second.

Equating the lunar and solar solutions for orbitals instead of the for energies, which we just did,

we have

Yields

25.

Where . The accuracy of this is

Thus the energy equations gave the equation 55:

16. !

And equating the orbital equations gives

26.

These last two yield

27.

The accuracy is

1secon d = d

⊙

2ℏ

⊙

⋅

⊙

⋅

⊙

ℏ

⊙

3/2 = cos(π /6)

(2.8314E 33)

(5.972E 24)

(7.34763E 22)

= (0.866)(9.13E 38)

(1737400)

(6.96E8)

4.29059E 72 = 4.5005E 72

4.29059E 72

4.5005E 72

100 = 95.3 %

⊙

⋅ ℏ

⊙

2 3

⋅

⊙

⋅ ℏ

⊙

= 1

of 21 64

Is 98% accuracy. The important thing that comes out of this is our base unit of a second because

we see it is also a function lunar, solar, and earth masses.

24.

(400.5986)

5.972E 24kg

1.9891E 30kg

= 1

0.98 = 1

1secon d = d

⊙

of 22 64

Part 3: The Characteristic Length For Habitable Planets

of 23 64

From part 1

We have

And, from Newtonian mechanics that

With these equations we can write:

Equation 24 in part 2 is

Using from Newtonian mechanics

equation 2 becomes

Where for the orbital number of the Earth

moon

ℏ

⊙

⋅

= 1.0secon d

ℏ

⊙

= (1secon d )K E

K E

ℏ

⊙

K E

⊙

ℏ

⊙

1secon d = 2r

⊙

K E

= n

⊙

4 c

⋅

n = 3

of 24 64

Let’s see just how accurate equation 3 is:

=(0.4330)(1,477m)(5.37068E5)=3.43477E8 meters

The orbital radius of the Moon today is 3.84E9 meters. The accuracy of our equation is

We have that the orbital radius of the moon is proportional to the cube of the mass of the planet

that it orbits to the cube of its mass. That is

We guess this is a condition for an optimally habitable planet, and that the other is:

Which is the condition for a perfect stellar eclipse by the moon of the planet. The term in the

equation

is easily recognizable. It is of the form of the Schwartzchild radius of a star, which is the radius

for which the star’s escape velocity is the speed of light, and will thus becomes a black hole:

Also is an important characteristic distance in cosmology. If is the mass of the

universe it is used to determine the curvature of the Universe and the expansion rate of the

Universe. Thus we have

Which we could propose is the characteristic length for a planet to be optimally habitable. We

could guess this means the planet would have a moon at .

moon

(

⊙

)

(0.4330)

(

(6.674E − 11)(1.989E30kg)

(299,792,458m /s

)

(5.972E 24)

(7.347E 22)

3.43477E8

3.84E8

100 = 89.447 % ≈ 90 %

∝

planet

moon

star

moon

(

⊙

4 c

)

= 369.25m eters

2GM

GM /c

star

4 c

moon

∝ M

planet

moon

of 25 64

Part 4: The Ancient System of Measuring Time and the Contemporary Second

of 26 64

It is worth looking at ancient systems of looking at time because that is where the origins of our

current systems began. Indeed our system came from the ancient Sumerians and Babylonians.

They divided the day into 12 units where the day is given by the rising and setting of the Sun,

which in turn is given by the period of the rotation of the Earth. Thus the day was given by the

period from the rising of the Sun to its setting, which was divided into 12 units which we call,

today, hours. Thus the day from sunrise to sunrise, or sunset to sunset, is 24 hours. Why they

chose 12 units could come from the fact that 12 is the smallest abundant number, which means it

has a lot of divisors: 1, 2, 3, 4, 6. Abundant means their sum is greater than 12 itself:

1+2+3+4+6=16. We know for certain they chose 12 because there are three sections on each

finger, so with 4 such fingers your can touch each such section with your thumb to count to

twelve. The Babylonians got base 60 from the Sumerians and further divided each hour into 60

minutes, and each minute into 60 seconds. Why base 60 was chosen is that it has a lot of

divisors as well, including the first 6 integers. It is the smallest number that does this.

We see dividing the day into 24 hours and dividing that further with base 60 lead to the duration

of a second we have today. Only on planet Earth would we have the primitive, ancient origins of

our mathematics in the end line up with modern physics in that, as it would turn out, it gave us

our basis unit of a second to be a natural constant. Though we could guess that on other planets

the ancient civilizations when first inventing mathematics and astronomy, would use base 60

because it is so convenient for doing math being evenly divisible by the first 6 integers. However

we did that and combined it with divisions of 24 units. Not necessarily would any planet do that.

However, we had other reasons to do that; the Moon orbits the Earth close to 12 times in the

time it takes the Earth to go once around the the Sun. However, there is an ancient system

where the people didn’t divide the day into 24 units, but rather into 60 units, meaning that their

hour was 24 minutes long. That is and

. This was the Vedic time-keeping system in ancient

India. Leave it to the Hindu Indians to have extraterrestrial intelligence in their ancient

beginnings. So let’s go into this. We will see that base 60 combined with 24 describes the

angular momentum of the Earth. That means it doesn’t just include the rotation period of the

Earth, but the size of the Earth (its radius) and the mass of the Earth.

Indeed if the dynamics of the factors the ancients gave us to create the duration of a second are

connected to the dynamics of stars systems then the second should define the rotational angular

momentum of the Earth since we divide the rotation period of the Earth into these factors to get

the unit of second, and should be connected to our Planck constant for the solar system which is

in units of angular momentum as well, as is Planck’s constant for the atom.

The angular momentum of the Earth with respect to the Sun is 2.66E40 kg m2/s. The rotational

angular momentum is 7.05E33 kg m2/s. In orbit angular momentum is given by

For a uniform rotational sphere it is given by

We found our solar system Planck constant was

This gives

(24h ours)(60mi n /h our) = 1440min /d a y

1440mi nutes/60 = 24minutes /h our

L = 2π M f r

L =

π M f r

ℏ

⊙

= 2.8314E 33J ⋅ s

of 27 64

We are now equipped to show that the ancient Sumerians were right in dividing the rotation

period of the Earth (the day) into 24 units (the hour) because

That is

Which is to say the angular momentum of the Earth to its Planck constant gives the base 60

counting in terms of the 24 hour day, the 60 of 60 minutes in an hour, and 60 seconds in a

minute, that determine our base unit of duration we call a second that happens to be, as I have

shown, the base unit of the wave solution to the atom and the Earth/Moon/Sun system.

Our equation in this paper for the Earth energy as a solution of the wave equation (eq. 13)

13.

does not depend on the Moon’s distance from the Earth, only its mass. The Moon slows the

Earth rotation and this in turn expands the Moon’s orbit, so it is getting larger, the Earth loses

energy to the Moon. The Earth day gets longer by 0.0067 hours per million years, and the

Moon’s orbit gets 3.78 cm larger per year. Equation 1

only specifies divide the day into 24 units, and hours into 60 minutes and minutes into 60

seconds, regardless of what the Earth rotational velocity is. But it was more or less the same as it

is now when the Sumerians started civilization. But it may be that it holds for when the Earth

day is such that when the Moon perfectly eclipses the Sun, which we said might be a condition

for the optimization of life preventing extreme hot and cold. That is when the following holds

Which holds for today and held for the ancient Sumerians and holds for when the Earth rotation

gives the duration of a second we have today.

We want the basis set of equations for the solar system. We have

We have from equation

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

2.5(24) = 60

earth

ℏ

⊙

24 = 60

K E

= n

⊙

⋅

2ℏ

⊙

earth

ℏ

⊙

24 = 60

planet

moon

≈

star

moon

ℏ

⊙

= 3.0281E8m

of 28 64

and,

From these it becomes clear that

Thus combining equation 2 with the following…

We write

From 3 we have

Which gives us

This is very accurate to give us a second. But notice 6.262 is approximately . We see that

moon

ℏ

⊙

⋅

= 1.0secon d s

ℏ

⊙

= (hC )K E

hC = 1secon d

ℏ

⊙

K E

ℏ

⊙

earth

ℏ

⊙

24 = 60

ℏ

⊙

= (1secon d )K E

ℏ

⊙

= (1secon d )

ℏ

⊙

ℏ

⊙

= (1secon d )GM

1secon d =

(

)

earth

(

)

(7.05E 33)

(6.67408E − 11)((7.347673E 22kg)

(299792458m /s)

(

)

6.262sec = 1.002 secon ds = 1.00secon d s

2π

of 29 64

is the circumference of a unit circle, it could be that the circumference of the Earth orbit, can

be taken as 1 (unity) and essentially we have the mystery of base 60 and the 24 hour day of the

ancient Sumerians is solved, it connects unit circle to 1 (unity).

The Hindu Vedic System has a day is 60 ghatika of 24 minutes each, each ghatika is divided

into 60 palas of 24 seconds each, and each pala is divided into 60 vispalas, each vispala of 0.4

seconds each. So where our system has a base unit of 1 second, theirs has a base unit of 0.4

seconds, so that could be an advantage to their system, a smaller unit of time is more reﬁned.

Further their day is divided into 60 units, ours into only 24, so their hour, the ghatika is only 24

minutes long, and ours is 60 minutes long, The proponents of this system in India say since

when working and doing chores we do a few chores in an hour, they do about one per ghatika

is 24 minutes, which makes the measure of time more manageable. That could be an

advantage I think. They say our hour is so long because the lines had to be far apart on the

Egyptian Sun Dial so the shadow cast by the Sun didn’t cross-over onto another line. But

today, with modern technology we can make clock lines marking hours closer together and

measure them with a pointer hand pointing to them without any problems, and the result is we

would have a smaller more reﬁned hour (60 of them in a day as opposed to 24). They suggest

this method of measuring time would work better in science and engineering as well, that we

have to get away from the way sun dials had to be made in Egypt in ancient times.!

But they further point out that their system describes Nature. They say theirs are 108,000

vispalas in a day, and 108,000 vispalas in a night giving 216,000 vispalas in a 24 hour day. The

diameter of the Sun is 108 that of the Earth, and the average distance from the Sun to the

Earth is 108 solar diameters, and the average distance from the Moon to Earth is 108 lunar

diameters. 108(10)(10)(10)=108,000. Ourselves and them use base 10 counting, and that is

probably because we have ten ﬁngers to count on.!

Conclusion

In conclusion we have that the unit of a second as derived from the Mesopotamians is a Natural

constant. As well we see the Hindu Vedic system describes Nature as well. It says

( )

1 =

(

)

2π

cos(45

∘

) = cos(π /4) =

⊙

= 2(6.957E8m) = 1.3914E 9m eters

⊙

= SolarRa diu s

= 1.496E11m

= Ear thOrbit

⊙

= 107.52 ≈ 108

moon

= 2(1.74E6m) = 3.48E6m

moon

= Lu n arRa diu s

moon

= 3.845E8m

moon

= Lu n ar Orbit

of 30 64

To see how this is connected to nature we would have to go into the significance of base 10

counting as we did with base 60 counting. But interesting here is that

because

13.

where

Which is to say radius of the Sun to the radius of the Moon in the planetary equation plays the

role of the number of protons (Z) in an element for the energy of the electron orbits in the

atomic equation and that the Moon perfectly eclipses the Sun as see from the earth (eq 15) might

be a condition for sophisticated habitable star systems, we know the Moon allows for the

seasons making life very successful on Earth by preventing temperature extremes. So in the

Hindu Vedic system the diameters of the Sun and Moon fitting into Earth orbit and Lunar orbit

the same amount might play a similar role in a theory. You would have to go into base 10 like we

did with base 60. We have…

moon

= 110 ≈ 108

(86,000sec /d a y)

43200

0.4sec /vispala

= 108,000vispala s /d a y

108(10)(10)(10) = 108,000

⊙

= 107.52 ≈ 108

moon

= 110 ≈ 108

⊙

6.957E8m

6,371,000m

= 109 ≈ 108

K E

= n

⊙

⋅

2ℏ

⊙

planet

moon

≈

star

moon

⊙

6.96E8m

1737400m

= 400.5986

of 31 64

West

24 hours

60 minutes

60 seconds

India

60 hours

24 minutes

24 seconds

A sort of inversion of things. The incredible thing here is that the wonderful equations:

unify the Mesopotamian system with the Hindu system because

This suggests that the Hindu system is just as Natural as the Western system.

Equation 6 that says

1 hour=(60min/hour)(60sec/min)=3600sec=1 Western Hour

1 vispala=0.4sec

60 vispala=(60)(0.4sec)=24 sec=1pala

60 pala = (60)(24 sec)=1440 sec=24 min = 1 ghaticka = 1 HinduHour

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

ℏ

⊙

24 = 60

1vi spal a = 0.4secon d s

(0.4s)(2.5) = 1secon d

1 =

(

)

2π

1 =

(1440sec)

(3600sec)

2π

1 =

(1gh a t ick a)

(1h our)

2π

1 =

Hin du Hour

Wester n Hour

2π

of 32 64

where is the circumference of a unit circle, and we have unity on the left, so it could be if you

combine the Indian system of measuring time with the Western system you would have a system

of units for the solar system where the equations work out nicely.

We have said the ancient Sumerians gave us the duration of a second we have today by

dividing up the rotation period of the Earth into 24 units we call hours, which the Babylonians

divided into 60 minutes and 60 seconds that they got from the Sumerian base 60. We have

found in my theory that the 1 second is a natural constant because it is the base unit of our

solution to both the solar system and the atom. We further found the Hindu’s used similar

numbers in an inverted way to the way of the West that came from the Sumerians, and that

their system relates not just to the West’s but to our theory in a dynamic way concerning the

angular momentum of the Earth and our Planck constant for the Solar system. !

The Sumerians of Mesopotamia and the Hindus of the Indus Valley Civilization were of the ﬁrst

civilizations, but so was Egypt. If the second was in Mesopotamia and in India, it was only

necessary to ﬁnd it in Egypt. Since the Earth spin has been the natural clock for which the

humans ﬁrst measured time since ancient times, the rising and setting of the Sun due to it, I

looked at how much distance through which the earth rotates at its equator in one second. It

is, since this distance is!

where theta is in radians, and the the radius of the Earth is 6.378E6m. The earth rotates

through 360 degrees per 24 hours, or per 86,400 seconds = 0.004167 degrees per second =

7.27E-5 radians. We have!

If the second was to exist in ancient times in Egypt we should see it in the Great Pyramids. The

three largest are in a line and are separated by!

Khufu to Khafre (0.25 miles)!

Khafre to Menkaure (0.3 miles)!

We see the second and third pyramids built, Khafre and Menkaure, which were built in the 25th

century BC and were of the the 4th dynasty of the old kingdom, are 0.3 miles apart, the

distance through which the Earth rotates in one second, the one second we found to be a

natural constant at the basis of the the atom and solar system.!

The ancient Sumerians say they got there mathematics in their writings from the Gods who

came to Earth from the sky they called the Anunaki. Some have suggested the Anunaki were

ancient aliens. The Ancient Egyptians built enormous pyramids from heavy stones weighing

tons, somehow lifted 50 feet to be stacked in near perfect mathematical proportions. This has

been a deep mystery in Archaeology. Again some suggest ancient aliens here. As well India has

1 =

(60pal a)

(60mi n)

2π =

(24min)

(60mi n)

2π

1 =

((60)(24sec))

((60)(60sec))

2π =

(60pal a)

(60sec)

2π =

(24sec)

(60sec)

2π

s = r θ

s = (6.378E6m)(7.27E − 5ra d ) = 464m = 0.464k m = 0.2883mi ≈ 0.3mi

of 33 64

its mandalas, visual painted patterns that are said to represent the sounds of chants to use to

take them into deep meditational states. The patterns found in their mandalas match the

patterns created by sound putting sand grains on a plate with transducers to vibrate them with

tones in modern acoustics studies. Again, some have suggested ancient aliens here. I have

shown all three of these civilizations have a system of measurement (the Egyptians had a 24

hour day as well) that is intrinsic to the laws of nature describing the atom and the solar system

in modern times. It can be suggested that the second was given to them by ancient aliens.

A really good ﬁnd is the following…!

of 34 64

My theory suggests the Sumerians could have ultimately got the unit of a second for

measuring time from ancient Aliens, they called the Anunaki who the say came from the sky.

The same second I have found is characteristic of our solar system and the proton. Scholars

have puzzled over the the depiction in Sumerian art of the strange watch-like bracelets around

the wrists of the Anunaki gods, who they say came from the sky and gave them mathematics.

They have twelve divisions like our clocks today have because the 12 and 6 o’clock positions

have two pointers running together as one.!

of 35 64

Part 5: Biological Life Part of a Universal Natural Process In The Universe

of 36 64

Introduction I have a theory for the star systems, the planets and their suns, wherein such

systems are solved with the Schrödinger wave equation that is used to solve atomic systems. The

result is that star systems and atomic systems, systems on the macro scales and micro scales, are

governed by the same underlying principles. It came to pass that this theory indicated an

overlap with biological systems indicating that biological life could be part of the same idea, and

that is what I hope to pull out of that paper and develop here.

The theory for the planets, which also predicted the radius of a proton, which is one of the

fundamental particles out of which matter is made, suggested that star systems which support

life are a part of a Universal natural process. One of the conditions for star systems that support

life made use of the interesting fact that has mystified science for some time now, that the Moon

is the right size and distance from the Earth that it nearly perfectly eclipses the Sun. The reason

a moon would be necessary for life to be optimally successful on a habitable planet is that we

know our moon that orbits our Earth, makes life here successful because it holds the Earth at its

inclination to its orbit around the Sun allowing for the seasons, thus preventing extreme hot and

extreme cold.

Interestingly, the solution for the planets of the Schrödinger wave equation, was quantized in

terms of the Earth’s moon, and a base unit of one second. The Moon seems to be some sort of a

natural yardstick. The strange thing is that the basis unit of time is one second and the second

was not developed with the planets and atoms in mind, so it is strange that it lands upon the

function of a natural constant in our theory. In the theory (An Abstract Theory of Reality,

Beardsley 2024) I talk a little about why that might be. The second came to us from the Ancient

Sumerians, who invented civilization with settling down from following the herds and hunting

with stone spearpoint to invent agriculture, metallurgy, writing, and mathematics. They had a

base 60 counting system, and found it convenient then to divide the Earth day (rotation period)

into 24 hours, which in the end became adopted by the world. Their base 60 counting resulted in

the Babylonians dividing the hour into 60 minutes, and that in turn into 60 seconds, and that is

how we have the duration of a second we have today. They chose base 60 (sexagesimal) because

60 is evenly divisible by so much from which they gave us the 12 hour day and 12 hour night or

24 hour day from sunrise to sunrise, sunset to sunset. 12 times 5 is 60, also 60 times 6 is 360,

they gave us the 360 degree circle as well.

In this paper I show that the same unit of a second that describes planetary systems and atomic

system in common describes hydrocarbons, the skeletons of biological life chemistry. I further

show that it predicts the atomic radii of the hydrogen and carbon atoms from which such

skeletons are made. Hydrogen and carbon are the most abundant elements in life chemistry,

carbon is the core element upon which life is made, and hydrogen is the simplest element,

element one in the periodic table of the elements, consisting of 1 proton, and is the most

abundant element in the universe by far, and plays the dominant role in all of chemistry. We will

see here that as much as the spacetime operators describe the solar system and the proton, they

describe the hydrocarbons as well, the skeleton of biological life chemistry. Further reason to

suggest life may be part of a Universal Natural Process.

of 37 64

A Theory For Biological Hydrocarbons I have found that the basis unit of one second is

not just a Natural constant for physical systems like the atom and the planets around the Sun

(See my paper An Abstract Theory of Reality, Beardsley 2024) but for the basis of biological

life, that it is in the sixfold Nature of the chemical skeletons from which life is built, the

hydrocarbons. I found

1).

2).

Where is the fine structure constant, is the radius of a proton, is the mass of a

proton, is Planck’s constant, is the constant of gravitation, is the speed of light, is the

kinetic energy of the Moon, is the kinetic energy of the Earth, and is the

rotation period of the Earth is one day.

The first can be written:

3).

4).

From which instead of saying the left sides of these equations are seconds, we say they are

proton-seconds by not letting the units of cancel with the bodies of these equations on the

left, which are in units of mass, but rather divide into them, giving us a number of protons. I say

this is the biological because these are the hydrocarbons the backbones of biological chemistry.

We see they display sixfold symmetry. I can generate integer numbers of protons from the time

values from these equations for all of the elements with a computer program. Some results are:

1secon d =

6α

⋅

4πh

1secon d =

K E

(Ear th Da y)

α = 1/137

K E

Ear th Da y

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon ds = h ydr ogen(H )

of 38 64

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of

the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons

the backbones of biological chemistry. And carbon is the core element of life. We see the

duration of the base unit of measuring time, 1 second, given to us by the ancients (the base 60,

sexagesimal, system of counting of the Sumerians who invented math and writing and started

civilization), is perfect for the mathematical formulation of life chemistry. Here is the code for

the program, it finds integer solutions for time values, incremented by the program at the

discretion of the user:

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792458,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

We have that

Since this 6 seconds is also proton-seconds we have

5). is carbon (C)

6). is hydrogen (H)

⋅

4πh

G c

= 18769

0.833E − 15

1.67262E − 27

4π (6.62607E − 34)

(6.67408E − 11)(299,792458)

= 6.029978s ≈ 6s

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

of 39 64

Our Theory For Hydrocarbons With all that has been said we are equipped to proceed. We

want to consider the radius of a hydrogen atom and the radius of a carbon atom. The radius of

a carbon atom given in your periodic table of the elements is often 70 to 76 picometers. The

covalent radius of hydrogen is given as 31 picometers. The atomic radius of hydrogen is 53

picometers and the atomic radius of carbon is 67 picometers. We want to consider the atomic

radii of both, because the covalent radius, determined by x-ray diﬀraction for diatomic

hydrogen, is the size of two hydrogen atoms joined H2 divided by two, where it is measured

that way, joined, in the laboratory. Carbon is C2 divided by 2. We are interested in the single

carbon and hydrogen atoms, because we want to know what our theory for their six-fold

symmetry with one another in their representations in proton-seconds says about the way they

combine as the skeletons of life chemistry. We start with the Planck constant, which is like

ﬂux, a mass (perhaps a number of particles) per second over an area. That is it is kilograms per

second over square area:!

We have equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

We can write these

3).

4). !

We have from 3 and 4…!

5). !

We ﬁnd the ratio between the surface areas of the hydrogen and carbon atoms:!

h = 6.62607E − 34J ⋅ s = 6.62607E − 37

⋅ m

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

(

6.62607E − 37

⋅ m

)

6secon d s

(

6.62607E − 37

⋅ m

)

6secon ds

1.67262E − 27kg

= 2.37689E − 6m

(

6.62607E − 37

⋅ m

)

1secon d

(

6.62607E − 37

⋅ m

)

1secon d

6(1.67262E − 27kg)

= 6.602486E − 8m

⋅

(6seconds)

⋅

(1second)

2.37689E − 6m

6.602486E − 8m

= 35.9999 ≈ 40

of 40 64

6). !

7). !

It is the golden mean. These atomic radii are the radii between the nucleus of the atoms and

their valence shell, which is what we want because the valence shell is the outermost electrons

responsible for the way the hydrogen and the carbon combine to make hydrocarbons. We will

write this!

8). !

I am guessing the reason we have the golden mean here is that it is the number used for

closest packing. But what we really want to do is look at the concept of action, for hydrogen

given by six seconds and carbon given by 1 second. We take equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

And, we write

9).

Where is the radius of the atom, and t its time values given here by equations 1 and 2. We have for

hydrogen

10).

This is actually very close to the radius of a hydrogen which can vary around this depending on how you

are looking at it, which we said is given by 5.3E-11m. For carbon we have:

11).

And this is actually very close to the radius of a carbon atom which is 6.7E-11m. The thing is if we consider

the bond length of the simplest hydrocarbon CH4, methane, which can be thought of as

surface

= 4π (r

) = 3.52989E − 20m

surface

= 4π (r

) = 5.64104E − 20m

4π(53pm)

4π(67pm)

= 0.62575 ≈ 0.618... = ϕ

= ϕ

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

G c

∫

dt = r

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s ) = 6.44E − 11m

of 41 64

16).

our, equations give

17).

which are the same thing. Thus we have the basis for a theory of everything in that it includes the macro

scale, the Earth/Moon/Sun System, because we had:

and it includes the radius of the proton which is the radius and mass of a proton and gives 1

second

and we have the hydrocarbons the skeleton of the chemistry of life give one second

1). is carbon (C)

2). is hydrogen (H)

and because they predict the radii of the carbon and hydrogen atoms at the core of life

10).

11).

+ r

= 5.3E − 11m + 6.7E − 11m = 1.2E − 11m

+ r

= 5.5E − 11m + 6.44E − 11m = 1.2E − 11m

1secon d =

K E

(Ear th Da y)

1secon d =

6α

⋅

4πh

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s ) = 6.44E − 11m

of 42 64

Part 6: Solutions For Jupiter And Saturn

of 43 64

Solutions For Jupiter And Saturn

Indeed, if we have the solution for the Earth/Moon/Sun system, and we want to say we have the

solution for the Solar System of the wave equation, we have to at least provide the solutions for

Jupiter and Saturn, which should suffice because they carry the majority of the mass of the Solar

System significantly. Since this paper mainly deals with the Earth/Moon/Sun system we will

presently leave out the solutions for the ice giants, Uranus and Neptune.

We have the solution for the Bohr hydrogen atom is

is the atomic number, the number of protons orbited, but for our system it is 1, because one

body is orbited. is the orbit number, which we will deal with later. We have our solution for the

Earth/Moon/Sun system is

Let is the earth mass, is the lunar mass, and is our

Planck constant for the Earth/Moon/Sun system. We compute equation 2:

The kinetic energy of the Earth is

This is

The Moon perfectly eclipses the Sun which means as seen from the Earth the Moon appears to

be the same size as the Sun. This is because

Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,

and is the Moon’s radius. This is because though the Moon is 400 times further from the Sun

than it is from the Earth, the Sun is 400 times larger than the Moon. That is

E = −

)

2ℏ

E =

⊙

M = M

m = M

⊙

= 2.8314E 33J ⋅ s

E =

(6.67408E − 11)

(5.972E 24kg)

(7.347673E 22kg)

2(2.8314E 33)

3.93E 30Joules

K E

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

1/n

= 697

≈

⊙

6.96E8m

1737400m

= 400.5986

of 44 64

Thus

We find the quantum mechanical solution to the Earth/Moon/System is

5 !

for Earth orbit (Third Planet).!

We see the solar Planck constant we developed works in solving the Schrödinger wave equation

for the Earth/Moon/Sun system. My belief as to why this was never done is that it had to be

realized that the solution of the Earth kinetic energy around the Sun as a quantum mechanical

state is based on the Moon around the Earth.

Now we want to find what the wave equation solutions are for Jupiter and Saturn because they

significantly carry the majority of the mass of the solar system and thus should embody most

clearly the dynamics of the wave solution to the Solar System.

I find that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid,

and go to the gas giants and ice giants, the atomic number is no longer squared and the square

root of the the orbital number moves from the numerator to the denominator. I believe this is

because the solar system here should be modeled in two parts, just as it is in theories of solar

system formation because there is a force other than just gravity of the Sun at work, which is the

radiation pressure of the Sun, which is what separates it into two parts, the terrestrial planets on

this side of the asteroid belt and the giants on the other side of the asteroid belt. The effect the

radiation pressure has is to blow the lighter elements out beyond the asteroid belt when the

solar system forms, which are gases such as hydrogen and helium, while the heavier elements

are too heavy to be blown out from the inside of the asteroid belt, allowing for the formation of

the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the atomic

number of the heavier metals such as calcium for the Earth, while the equation for the giants has

the atomic numbers of the gasses. We write for these planets

So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):

⊙

= 400

E = n

⊙

⋅

⊙

n = 3

3(400.5986)(3.93E30J ) = 2.7268585E 33J

2.7268585E33

2.7396E33

= 99.5 %

E =

2ℏ

⊙

K E

(1.89813E 27kg)(13720m /s)

= 1.7865E 35J

of 45 64

Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that

Our equation for Jupiter is

Where is the atomic number of hydrogen which is 1 proton, and for the orbital

number of Jupiter, . Now we move on to Saturn…

The equation for Saturn is then

It makes sense that Saturn would use Helium in the equation because Saturn is the next planet

after Jupiter and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well,

just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.

Our equation in this paper for the Earth orbit does not depend on the Moon’s distance from the

Earth, only its mass. The Moon slows the Earth rotation and this in turn expands the Moon’s

orbit, so it is getting larger, the Earth loses energy to the Moon. The Earth day gets longer by

0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year. It is believed

the Moon came from a chunk knocked off the Earth from a collision with a Mars sized

E =

(6.67408E − 11)

(1.89813E 27kg)

(7.347673E 22kg)

2(2.8314E 33)

E =

(3.971E 35J ) = Z

(1.776E 35J )

1.7865E 35J

1.776E 35J

= 1.006proton s ≈ 1.0proton s = hydrogen(H )

Z = Z

E =

2ℏ

⊙

n = 5

K E

(5.683E 26kg)(10140m /s)

= 2.92E 34J

E =

(6.67408E − 11)

(5.683E 26kg)

(7.347673E 22)

2(2.8314E 33)

2.45

(3.5588E 34J ) = Z(1.45259E 34J )

Z(1.45259E 34J ) = (2.92E 34J )

Z = 2proton s = Heliu m(He)

E =

2ℏ

⊙

of 46 64

protoplanet. At the time that the Moon formed it was about 4 Earth radii distant 4.5 billion

years ago. After 100 million years the Moon became tidally locked making its rotation period

equal to its orbital period keeping one face always towards the Earth. After 500 million years the

Moon was orbiting at about 20 Earth radii. It is believed that during the period of heavy

bombardment in a chaotic early solar system about 4.1 to 3.8 billion years ago a large object did

a close pass pulling the Moon further changing its orbit giving it its 5 degree offset from the

Earth’s equator. Our wave equation solution may only use the Moon’s mass but the equation for

kinetic energy of the Moon to kinetic energy of the Earth times the Earth Day equal to about one

second:

Which we connect with the equation where the proton gives one second

This holds for when the Moon was at a distance from the Earth such that it appears the same

size as the Sun, which means:

Which is when the two equations above for one second connect to our wave equation solution to

the Earth

Because .

The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the

Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get too

extremely hot or too extremely cold. We see the Moon may be there for a reason.

K E

moon

K E

earth

(Ear th Da y) = 1.08secon ds

6α

h 4π r

= 1.004996354secon ds

planet

moon

≈

star

moon

K E

= n

⊙

⋅

2ℏ

⊙

star

moon

⊙

of 47 64

Part 7: Modeling Habitable Star Systems

of 48 64

Modeling Habitable Star Systems

We want to solve a habitable planetary system in general and apply it to another star system. We

must include the solution of its moon as well, because the Moon of the Earth makes life

optimally possible. However, our Moon is counter-intuitive, it breaks all of the rules of the other

moons in our solar system. One is it has a low mass for its size. It has been said that it could be

hollow. Even suggested that it is a hollow-spacecraft put there to make life as we know it

possible. In our quantum theory for the solar system, we may have discovered the science of the

engineers who may have made the Moon, and we are possibly applying their science here. Our

quantum theory for the solar system also has a solar solution as compared to a lunar solution, so

it works for star systems in general.

The perfect star system to use for our purposes here would be Tau Ceti. Tau Ceti is a star

spectrally similar to the Sun and the closest of such a G-class star, at about 12 light years distant.

Its mass is 78% of the Sun’s. Its mass is

Its radius is:

And its luminosity is

It has several unconfirmed planets with one potentially in the habitable zone. It has been the

subject of science fiction literature. It is in the constellation Cetus. From Tau Ceti the Earth

would be seen to be in the northern hemisphere constellation Bootes with an apparent

magnitude of 2.6. It is spectral class G8 V where our Sun is G2 V.

For the luminosity of a main sequence star we only need to know the mass of the star to get its

luminosity:

This holds in solar masses and solar luminosities. The habitable zone of a star is given by

luminosity. If the star is 100 times brighter than our Sun, then the habitable zone is 10 times

further than the Earth is from the Sun by the inverse square law.

In astronomical units and solar luminosities. From we can get the orbital period in Earth

years:

We can get the radius of the Moon of the planet if we know the mass of the planet, :

= 0.783 + / − 0.012M

⊙

= 0.793 + / − 0.004R

⊙

= 0.488 + / − 0.0010L

⊙

L = M

3.5

= L

= r

of 49 64

We want the mass of the Earth from the mass of the Sun. We would guess we can get it from the

mass of the Sun, the size of the Sun, and the base 60 sexagesimal system upon which we have

found everything is based. I find the relationship exists for the Earth:

(36)(5)=180=360/2 and 360/6=60 is base 60 sexagesimal. This gives

This is an accuracy of

We can get the orbital distance of the planet’s Moon from the planet:

Once we have that we can get the radius of the Moon of the planet

Now we can get the rotation period of the planet, its length of day:

Because the orbital velocity of the planet’s Moon is

We can use the the delocalization time get the moon’s mass, which is 1/2 the planet’s year, with

10 , See Appendix 1

⊙

= 1

⊙

= 0.13889(1.9891E 30kg)

6.96E8m

1.496E11m

= 5.979748E 24kg

5.972E 24

5.979748E 24

100 = 99.87 %

planet

moon

≈

star

moon

⊙

= 1

Ear th Da y =

⊙

τ =

moon

(2r

moon

)

ℏ

of 50 64

And the whole system is solved. Let us see how the theoretical luminosity appears with that

measured with equation 8.1:

It is close to the measured. The equation for predicting is approximate as the actual

values can vary a little with things like metallicity of the stars. Tau Ceti has low metallicity

compared to our Sun.

Let’s find the orbital distance of the habitable planet if it exists with equation 2:

Let’s find the orbital period of the habitable planet (It’s year) with equation 3:

Let’s get the mass of the habitable planet from equation 5:

Which is good because the planet, if it exists, is thought to be larger than the Earth. Let’s get the

radius of its moon from equation 4:

Let’s see at what distance it orbits the planet with equation 6:

= M

3.5

= 0.783

3.5

= 0.425L

⊙

0.448L

⊙

= 0.488 = 0.69857AU = (0.6986AU )(1.496E11m /AU ) = 1.045E11m eters

= r

= (0.69867AU )

= 0.584years(365.25d a ys)((24hrs)(60min)(60sec) = 1849638secon ds

(0.584)(365.25d a ys) = 213.306d ays

= (0.783)(1.9891E 30kg) = 1.5575E 30kg

= (0.793)(6.96E8m) = 5.52 E 8m

(1.5575E 30kg)

(5.52E8m)

(1.045E11m)

= 6.036E 24kg

6.036E 24

5.972E 24

= 1.0107Ear th Ma sses

= 2R

2(5.52E 8m)

(6.036E 24kg)

(1.5575E 30kg)

= 1.53656E6m =

1.53656E6m

1.7374E6m

= 0.8844Lun arRa dii

of 51 64

Let’s get the orbital velocity of this moon with equation 9:

So the orbital period of the moon is:

Compared to that of the Earth which is 27.3 days for the sidereal month and is a 29.53day lunar

month. The Moon of Tau Ceti would have a slightly longer lunar month, too. Now we can

determine the rotation period of the planet which would be its day:

We get the mass of the moon from equation 10:

So the rotation period of the planet (Its Day) is from equation 8:

If the moon of Tau Ceti is to perfectly eclipse its star, like our Moon does with the Sun, to let its

inhabitants know that they are there for reason, and so as to theoretically be a condition for

optimizing life, then we have the following should hold:

moon

= r

planet

moon

star

= (1.045E11m)

1.53656E6m

5.52E8m

= 2.91E8m

2.91E8m

3.84E8m

= 0.7578L un arOrbital Ra dii

(6.67408E − 11)(6.0636E 24kg

2.91E8m

= 1,176.6m /s

2π r

2π (.91E8m)

1,176.6m /s

= 1.554E6secon d s

(1.554E6s)

min

60sec

h our

60mi n

d a y

24h ours

= 17.987d a ys

ℏ

(2.8314E 33J ⋅ s)(0.5)(18429638s)

4(2.91E 8m)

= 7.7E 22kg

7.7E 22kg

7.34763E 22kg

= 1.048Ear th Moon s

Pl a n et Da y =

2(1.045E11m)

1,176.6m /s

7.7E 22kg

1.5575E 30kg

= 39495.61secon d s = 658.26min = 10.971h ours

10.971

24h ours

= 0.457Ear th Da ys

planet

moon

≈

star

moon

of 52 64

We have

And it does hold. For the Earth/Moon/Sun system we get

For the Tau Ceti system the 359 is close to 360, which is a convenient amount of degrees into

which divide a circle, like we did. We see the Tau Cetians might do the same, because like us

they might choose the base 60 sexagesimal system of counting, and this 360 might influence the

way the inhabitants of a planet orbiting this star would make their calendar. Just like the 365

day year here on Earth corresponds closely to the 360 degrees of a circle. Aside from having here

on Earth the degree, we have gradians, of which there are 400 in a circle. They make it very easy

for computing right angles to a given angle. They are used more in Europe and in surveying and

architecture, though it might have some interesting connection with our 400=400.

The habitable planet of Tau Ceti would orbit it star with a period of 213 days, which would be its

year. It star is about 78% the mass of our Sun, and about 79% its size. Its habitable planet would

be a little larger than the Earth, but not much. Its Moon would be about 88% the size of our

moon, but a little bit more massive, but not much more massive. It would orbit Tau Ceti once

about every 18 days meaning there would be 11.833 months in the Tau Ceti Year, or about 12

months like we have. It would orbit the planet at about three quarters the distance ours does.

The day from sunrise to sunrise, or sunset to sunset would be about 11 hours, or close to half the

length of our day. This is if Tau Ceti was engineered for life, but it doesn’t have to have a planet

so optimally designed for life as ours. Our 24 hour day would be better for when entering the

realm of doing astronomy, because longer nights mean more complete observations of the

universe surrounding us than for the Tau Cetians. It may mean for the conditions for life to be

optimal, like here on Earth, the star would have to be as massive as the Sun, to provide a longer

year and a longer day. But we see here how the science of engineering planetary systems optimal

for life, might be done.

1.045E11m

2.91E8m

5.52E8m

1.53656E6m

359 = 359

400 = 400

of 53 64

Part 8: The Meter A Natural Unit!

of 54 64

With the second as the characteristic time of the Universe, one might by tempted to look for

the corresponding length and mass units in the International System of Units commonly known

as the metric system, as Natural as well. Indeed I ﬁnd the meter presents itself nicely as a

candidate.!

The period of a pendulum depends on the acceleration of gravity at the surface of planet where

it is oscillating. On the Earth this acceleration is g = 9.807 m/s2. And, the period of the

pendulum, T, is related to the length of the cord, , or arm, to which the mass is attached that

swings at its end, as well. It has a period given by:!

We have said the second is the characteristic time of the Universe, and have said it is deﬁned

by the rotation period of the Earth and the sexagesimal mathematics given to us by the ancient

Sumerians. If the second comes to us from the rotational frequency of the earth, which is our

base unit of time, is it possible that our base unit of length in the metric system, the meter,

could come to us from The Earth’s gravity? As it would turn out, since the period of a

pendulum is given by a swing left and swing right again, that if the length of the pendulum is

one meter, that its period is 2 seconds, which means a swing left and a swing right takes about

one second. In other words, if , then then for a period of 2 seconds we have a length

because g=9.807m/s2 is approximately pi squared. It was Christian Huygens, a

Dutch mathematician and astronomer who ﬁrst noticed this in the 17th century. He suggested

we deﬁne the meter so that pi squared is exactly g, which would be very close to the meter we

have today. Let’s see this:!

Thus we have!

Since the gravity at the surface of the Earth is!

We have!

1). !

T = 2π

g = π

l = 1meter

T = 2π

= 2π

1meter

9.807m /s

= 2.006373secon ds

1.0secon ds = π

1meter

g =

1secon d = π

(1meter)

⋅

of 55 64

= !

We have equation 19 in part 1 is!

19)

And, equation 20 in part 1 is

20).

Which shows the to rotational angular momentum of the Earth is convenient for the ancient

Sumerian 24 hour day, 60 minute hour, and 60 second minute because the Solar System is

quantized in terms of . So, not only is the second defined in terms of the rotational frequency

of the Earth, but in terms of the mass of the Earth, that mass that is rotating, and the radius (or

size) of the Earth mass that is rotating. Considering our equation for the meter (equation 1):

1). !

We see it speaks of the radius of the Earth and the mass of the Earth, as well. The average

rotational angular momentum of the Earth is!

2). !

We have from 1!

3). !

We notice since equation 19 of part 1 is!

And since , angular velocity, that from equation 2:!

4). !

1meter(6.378E6m)

(6.67408E − 11)(5.972E 24kg)

= 1.0036secon d = 1.0secon d s

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

ℏ

⊙

24 = 60

ℏ

⊙

1secon d = π

(1meter)

⋅

earth

πM

86400s

1meter =

(1sec)

⋅ G ⋅

earth

ℏ

⊙

= 2.5

= 2π f

earth

ℏ

⊙

earth

of 56 64

Equation 4 is!

5). !

From equation 3!

3). !

We can make an equation 6:!

6). !

Putting equation 6 in equation 5 gives!

7). !

We have!

8). !

Where is the cross-section of the Earth, the disc of the Earth. This is an actual thing

in physics, in climate science, where the radiation of the Sun intercepts the Earth disc to

compute the average annual temperature of the Earth. It is also what we see visually when we

look at the Earth, say from the Moon, or as it would appear from Mars. This is good because a

big part of this paper is the apparent, what the ancients observed, like the Moon perfectly

eclipsing the Sun as seen from the Earth, which plays a role in our equation for the wave

equation solution of the Earth/Moon/Sun system, because the phenomenon has an algebraic

condition. We can write 8 as!

9). !

Since from equation 20 Part 1, equation 8 can be written:!

earth

= ℏ

⊙

earth

= ℏ

⊙

1meter =

(1sec)

⋅ G ⋅

= (1meter)

1sec

⋅

earth

= R

(

1secon d

)

ℏ

⊙

(1meter)

1secon d =

(

πR

earth

)

ℏ

⊙

(1meter)

πR

= A

1secon d =

(

earth

)

ℏ

⊙

(1meter)

earth

= ℏ

⊙

(

)

of 57 64

10). !

= !

too check our work so far and make sure we got the right units. Let’s check if the numbers

work, and if so, how well:!

, !

= !

And it works very well. Considering equation 10, since from equation 25, part 1:!

25)

We have

And 10 becomes:

11).

If we invert 10 we have it is acceleration:

12).

Again, since , and we have:

13).

(1secon d )

(1meter)

= π

(

)

(

ℏ

⊙

)

= A

(

)

(

ℏ

⊙

)

(

)(

kg ⋅ m

)

(

)

ℏ

⊙

= 2.8314E33

= 2π /86400sec = 7.2722E − 5sec

−1

= πR

= π(6.378E6E6m)

= 1.278E14m

G = 6.674E − 11

(1.278E14m

)

(

)

(7.2722E − 5)

(2.8314E33)(6.674E − 11)

= 1.00568

1 =

(

)

2π

(

)

(1secon d )

(1m eter)

2π

(

ℏ

⊙

)

(

)

(

ℏ

⊙

)

2π

(

)

= 2π f

(

ℏ

⊙

)

of 58 64

Since

And

We have

14).

So we have unit acceleration in terms of the Earth day. This is good, it hearkens back to one of

our three spacetime operators that solve everything, equation 2 in Part 1, which is

Which can equivalently can be written:

But we want that the meter itself is natural, as well. We have the second is the characteristic

time of the Universe. If the meter is natural, then the kilogram might be natural because the

density of water is is defined to be that a kilogram of water occupies a cube 10cm on each side.

But, then we would have to get into base 10 and the metric system. But we can show the meter is

natural. Equation 12 is

12).

And the ground state for the solar system in Part 1 is

Thus, the meter is

15).

86400s

= Ear th Rotation Frequ enc y = f

= Ear th Da y

(

ℏ

⊙

)

(Ear th Da y)

(2.8314E 33)(6.674E − 11)

(1.278E14)

86,400s = 0.9963

≈ 1

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

1secon d =

K E

(Ear th Da y)

(

)

(

ℏ

⊙

)

moon

ℏ

⊙

⋅

= 1.0secon d s

1m eter =

(

ℏ

⊙

⋅

)(

)

(

)

of 59 64

= =

X =

We have that it works very well. How could the meter connect with ancient aliens? The answer is

there are some who believe that the meter, and indeed the kilogram, were discovered as natural

by the ancient Egyptians, and that it worked its way into Europe before the metric system was

founded in France in the 1795. That the ancient Egyptians had had it encoded in their unit of

length, the royal cubit, by way of the volume of a sphere inscribed in a cube. This was presented

in the movie “Great Pyramid K 2019” by the director Fehmi Krasniqi. It would seem according

to Quentin Leplat, an independent researcher in ancient science, twelfth century Roman

churches may have used the metric system. He took measurements and did a statistical

treatment of the data. He found these monuments used what seemed to be both the royal cubit

and the meter. He found the Roman Church of Saint Nectaire had the entrance double of the

first entrance, which was 3 meters wide. And, that the entrance behind was 4 royal cubits wide

and that the length of this Roman church was exactly 40 meters. He found The Church of the

Conques had a stone that was different than all the others and that was much larger and that

this stone measured exactly 1 meter wide.

There is an interesting thing that happens with all of this. We have said the Moon is some kind

of a natural yardstick, it describes the Sun as having a radius of 400 in our energy solution of the

wave equation, and its mass determines the characteristic time of one second in the ground state

expression for the solar system. Now that we have unit acceleration, equation 14:

we see this is good to have because the gravity at the lunar surface is 1.62 m/s2. Since 1.62~1.618

which is the golden ratio, , we have

16).

(

ℏ

⊙

)(

ℏ

⊙

)

(

ℏ

⊙

)

(

)

(

)

(

)

(

(2.8314E 33)

6.674E − 11

)(

(2.8314E 33)

(7.3475731E 22)

)(

(2.8414E 33)

7.3476731E 22)

)

(

(1.278E14)

(7.2722E − 5)

)(

)

(

299792458

)

(1.201188E 77)(0.02020938)(7.1375925E − 36)(8.419232E − 25)(6.25)(8.97744E16)

1.01444m eters ≈ 1.0meters

(

ℏ

⊙

)

(Ear th Da y)

moon

= Φ

(

ℏ

⊙

)

(Ear th Da y)

(2.8314E 33))(6.674E − 11)

(1.278E14)

86400 = 0.99963m /s

≈ 1m /s

of 60 64

Also, one of the obvious things we could have done at the beginning of this section is to take

equation 3

And, the equation for the ground state of the Solar System

and write

17).

1m eter =

(1sec)

⋅ G ⋅

moon

ℏ

⊙

⋅

= 1.0secon d s

1m eter =

ℏ

⊙

⋅

(2.8314E 33)

(7.3474731E 22)

⋅

299792458

⋅

5.972E 24

(6.378E6)

= 1.00277m = 1.00m eters

of 61 64

Appendix 1

If we want to prove that our planetary Planck constant is correct, the delocalization time for the

Earth should be 6 months using it, the time for the Earth to travel the width of its orbit. We want

to solve the Schrödinger wave equation for a wave packet and use the most basic thing we can

which is a Gaussian distribution. We want to then substitute for Planck’s constant that is used

for quanta and atoms our Planck-type constant (h bar solar) for the Earth/Moon/Sun system

then apply it to predict the delocalization time for the Moon in its orbit with the Earth around

the Sun.

We consider a Gaussian wave-packet at t=0:

We say that is the delocalization length and decompose the wave packet with a Fourier

transform:

is the harmonics of the wave function. We use the identity that gives the integral of a

quadratic:

Solve the equation

With the initial condition

A plane wave is the solution:

Where,

The wave-packet evolves with time as

Calculate the Gaussian integral of

ℏ

⊙

ψ (x,0) = Ae

−

ψ (x,0) = Ae

−

∫

2π ℏ

ℏ

∫

∞

−∞

−α

x+βx

d x =

4α

iℏ∂

ψ (x, t) =



ψ (x, t)

ψ (x,0) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−ϵ( p)t)

ϵ( p) =

ψ (x, t) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−

of 62 64

and

The solution is:

where

is the delocalization distance, which for instance could be the width of an atom. is the

delocalization time, the average time for say an electron to traverse the diameter of the atom and

even leave it, to delocalize. If we substitute for our , and say that the delocalization distance

is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization

time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We

have

Where is the mass of the Moon, and is the orbital radius of the Moon. We have

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and Moon

travel from one position to the opposite side of the Sun. The closeness is

Thus we know our is accurate, it continues to function in a theoretical framework. The thing

about this is that it means we can predict the mass of the Moon from the Earth year. In terms of

what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at

its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a

mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for

water to exist as liquid, and thus could be as it is for a reason, which means life might be part of

a physical process throughout the Universe, that it unfolds naturally in the evolution of star

systems.

2ℏ

2mℏ

β =

i x

ℏ

= ex p

[

−

⋅

1 + t

/τ

]

τ =

m d

ℏ

⊙

τ =

moon

(2r

moon

)

ℏ

⊙

moon

τ = 4

(7.34767E 22kg)(3.844E 8m)

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

ℏ

⊙

of 63 64

Appendix 2: The Data For Verifying The Equations

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and

Earth’s orbital velocity at perihelion we have:

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G: 6.67408 × 10

−11

c : 299,792,458m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

K E

moon

(7.347673E 22kg)(966m /s)

= 3.428E 28J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

of 64 64

The Author!